Answer:
A. 0.025inches= 1m
So 75in= 0.025*75= 1.88m
B 3.45*10^6yrs= 3.45E6* 365x 86900s
= 1.09*10^8s
C. 62ft/day= 62* 0.3048/86900= 2.1*10^-4
D. 2.2*10^4mi² x ( 1609.3)²
= 6.1*10^7m²
A car travels 60 miles due West first then turns back and travels 120 miles due East in 3 hours. What is...
a) the distance traveled by car?
b) the displacement of the car?
Answer:
A. 180 milesB. 60 milesExplanation:
In this problem, we are required to solve for the total distance that the car travelled. and the displacement
A) the distance travelled by car
this can be gotten by summing all the distances the car has travelled.
i,e total distance= 60 miles+120 miles
total distance= 180 miles
B) the displacement of the car
the displacement can be gotten by subtracting the final distance from the initial distance
final distance = 120 miles
initial distance= 60 miles
displacement= 120-60= 60 miles
The table shows chronological events in the life of our Sun, a medium-sized star. Place the missing events into the correct rows of the table.
Hello, I don't see a table, but I am guessing that you are referring to the one I attached (below).
Answer:
So, the correct order of events sorted chronologically is:
1. A nebula located in the Milky Way galaxy begins pulling nearby hydrogen atoms in its orbit.
2. The Nebula shrinks in its volume due to gravity, becoming denser and hotter. But, it's not hot enough for nuclear fusion.
3. The temperature in the core of the Nebula reaches 14 million Kelvin.
4. Hydrogen atoms begin shedding their electrons and colliding with one another.
5. The Sun enters the main sequence stage. The energy created as a result of its radiation begins nurturing life on planet such as Earth.
6. The Sun uses up all the hydrogen in its core.
7. The Sun expands greatly and cools. It is larger and redder.
8. The Sun completely runs out of hydrogen to fuse. Its outer layers are pushed away, and a cloud of ionized gas surrounds its core.
9. The Sun is a white dwarf with a dim glow.
Hope this helps!
How much can having a lighted candle increase the temperature inside the shelter?
A lighted candle can increase the temperature inside a shelter by a few degrees Celsius.
A lighted candle produces heat by burning fuel. The heat from the candle is transferred to the air in the shelter by conduction, convection, and radiation. Conduction is the transfer of heat through direct contact. Convection is the transfer of heat through the movement of fluids. Radiation is the transfer of heat through electromagnetic waves.
The amount of heat that is transferred to the air in the shelter depends on the size of the candle, the number of candles, and the insulation of the shelter. The larger the candle, the more heat it will produce. The more candles, the more heat will be produced. The more insulation the shelter has, the less heat will be lost to the outside. In general, a lighted candle can increase the temperature inside a shelter by a few degrees Celsius.
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You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment building. You travel in the elevator 22.0 m upward to your floor, walk 12.0 m north to the door of your apartment, and then walk 6.0 m due east to your balcony that overlooks the street. Your friend is standing where you left her. how far are you from your friend?
Answer:
The answer is "26.3 m".
Explanation:
The positive value from the x-axis is to the direction east side
The negative value from the x-axis is to the direction west side
The positive value from the y-axis is to the direction upwards side
The negative value from the y-axis is to the direction down words side
The positive value from the z-axis is to the direction southside
The negative value from the z-axis is to the direction north side
If the value is i, j, and k are the unit of the given vectors, which can be defined as follows:
[tex]\hat i, \hat -i, \hat j, \hat -j, \hat k, \hat -k[/tex]
The displacements values:
[tex]\underset{d_1}{\rightarrow} = -14 \ \hat i\\\\\underset{d_2}{\rightarrow} = 22\ \hat j\\\\\underset{d_3}{\rightarrow} = -12 \ \hat k\\\\\underset{d_4}{\rightarrow} = 6 \ \hat i\\\\[/tex]
calculating the final displacement that is [tex]\underset{d_5}{\rightarrow}[/tex]:
[tex]\Rightarrow \underset{d_1}{\rightarrow}+\underset{d_2}{\rightarrow}+\underset{d_3}{\rightarrow}+\underset{d_4}{\rightarrow}+\underset{d_5}{\rightarrow} =0\\\\\Rightarrow \underset{d_5}{\rightarrow} \ \ = -(\underset{d_1}{\rightarrow}+\underset{d_2}{\rightarrow}+\underset{d_3}{\rightarrow}+\underset{d_4}{\rightarrow})[/tex]
[tex]=- (-14 \hat i+ 22 \hat j -12 \hat k+ 6 \hat i)\\\\=- (-8 \hat i+ 22 \hat j -12 \hat k)\\\\=8 \hat i- 22 \hat j +12 \hat k\\[/tex]
[tex]|\underset{d_5}{\rightarrow}|=\sqrt{8^2 +(- 22)^2 +(12)^2 }\\\\[/tex]
[tex]= \sqrt{64 + 484 +144 }\\\\= \sqrt{208 + 484}\\\\= \sqrt{692}\\\\= 26.3 \ m[/tex]
A 0.0125 kg bullet strikes a 0.240 kg block attached to a fixed horizontal spring whose spring constant is 2.25*10^3N/m and sets it into oscillation with amplitude of 12.4 cm. What was the initial speed of the bullet if the two objects move together after impact?
Answer:
The value is [tex]u_1 = 236 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of bullet is [tex]m_b = 0.0125 \ kg[/tex]
The mass of the block is [tex]M_B = 0.240 \ kg[/tex]
The spring constant is [tex]k = 2.25*10^{3} \ N/m[/tex]
The amplitude is [tex]A= 12.4 \ cm = 0.124 \ m[/tex]
Generally according to the conservation of momentum is
[tex]m_b u_1 + M_B u_2 = (m_b + M_B) v[/tex]
given that the block was at rest we have that
[tex]m_b u_1 = (m_b + M_B) v[/tex]
Now the angular velocity of the both bodies is mathematically represented as
[tex]w = \sqrt{\frac{k}{M_B + m_b} }[/tex]
[tex]w = \sqrt{\frac{ 2.25*10^{3}}{ 0.0125 + 0.240 } }[/tex]
[tex]w = 94.4 \ rad/s[/tex]
Given that the system after collision set into oscillation
The maximum linear velocity of the system after impact is mathematically represented as
[tex]v = A * w[/tex]
[tex]v = 94.4 *0.124[/tex]
[tex]v = 11.7 \ m/s[/tex]
From above equation
[tex]u_1 = \frac{(m_b + M_B ) * v}{m_b}[/tex]
=> [tex]u_1 = \frac{(0.240 + 0.0125) * 11.7}{ 0.0125}[/tex]
=> [tex]u_1 = 236 \ m/s[/tex]
An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spilled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of mercury in the open arm is 900mm above the lowest part of the manometer.
(a) When the gas is not flowing, the pressure is the same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connect to the pipe?
(b) When gas is flowing, the mercury level in the visible arm drops by 25mm. What is the gas pressure (psig) at thismoment?
Answer:
A. Using
Pgauge= Pmanometer
And we know that
Pgauge= deta(hpg)
So deta h = Pgauge/density x g
So
= 7.5(6894.76/1psig)/ 13.6*10^3*9.8m/s²)
= 387.9mm
So to find height of pipe connected to the pipe we say
= h -deta h
= 900-387.97mm
=512.02mm
B. We use manometry principle
Pgas+density xg(25*10^3)-PX density{h-(H-0.25)=0
So
Finally Pgas= 6.54psig
unit of speed is derived unit..why??
Answer:
[tex] \boxed{ \sf{see \: below}}[/tex]
Explanation:
The unit of speed is metre per second or m / s. It clearly depends on two fundamental units ; unit of length ( metre ) and unit of time ( second ). Hence , m/s is a derived unit.
Hope I helped!
Best regards! :D
The unit of speed is m/s. The two terms metre and second are fundamental quantities.
As,m/s is derived from these two fundamental quantities. Therefore, Unit of speed is a derived unit.
Two parallel wires I and II that are near each other carry currents i and 3i both in the same direction. Compare the forces that the two wires exert on each other. A. The wires exert equal magnitude attractive forces on each other. B. Wire I exerts a stronger force on wire II than II exerts on I.C. Wire II exerts a stronger force on wire I than I exerts on II. D. The wires exert equal magnitude repulsive forces on each other. E. The wires exert no forces on each other.
Answer:
A. The wires exert equal magnitude attractive forces on each other.
Explanation:
Magnetic field due to current i on current 2i
B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires
Force on wire II due to wire I per unit length
= magnetic field x current in wire II
= B₁ x 2 i
= [ 10⁻⁷ x 2 i / r ] x 2i
= 4 x 10⁻⁷ i² / r
Magnetic field due to current 2i on current i
B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires
Force on wire I due to wire II per unit length
= magnetic field x current in wire I
= B₂ x i
= [ 10⁻⁷ x 4 i / r ] x i
= 4 x 10⁻⁷ i² / r
So final forces on each wire are same .
This force will be attractive in nature . The direction of force can be known from fleming's right hand rule .
The meter was originally defined so that the period of a meter-long simple pendulum would be exactly 2.00 second. (Was your measured value close to this?) Given the relationship, T^2 alpha L (T^2 is proportional to L) what would be the length of a simple pendulum, in centimeters, with a period of exactly one second?
Explanation:
Given that,
Initial length of simple pendulum, [tex]L_1=1\ m[/tex]
Initial time period, [tex]T_1=2\ s[/tex]
We need to find the length of the simple pendulum when the period is exactly 1 second.
[tex]T_2=1\ s[/tex]
We know that the time period of simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{L}{g}} \\\\T\propto \sqrt{L} \\\\\dfrac{T_1}{T_2}=\dfrac{L_1}{L_2}[/tex]
Put all values and find L₂
[tex]\dfrac{T_1}{T_2}=\sqrt{\dfrac{L_1}{L_2}}\\\\L_2=\dfrac{T_2^2L_1}{T_1^2}\\\\L_2=\dfrac{1^2\times 100\ cm}{2^2\ s}\\\\L_2=25\ cm[/tex]
So, the length of the pendulum with a period of exactly one second is 25 cm.
A cat had climbed at the top of the top of a tree. The tree is 20 meters high and the cat was 6kg. How much potential energy does the cat have?
Please hurry and thanks!
Answer:
About 1200J
(If we take gravity to be 10m/s^2)
Explanation:
U=mgh
m=6kg
g=10m/s^2
h=20m
U=(6)(10(20)=1200J
Suppose A=BnCm, where A has dimensions LT, B has dimensions L2T-1, and C has dimensions LT2. Then the exponents n and m have the values?A. 2; 3B. 2/3; 1/3C. 4/5; -1/5D. 1/2; 1/2E. 1/5; 3/5
Explanation:
The expression is :
[tex]A=B^nC^m[/tex]
A =[LT], B=[L²T⁻¹], C=[LT²]
Using dimensional of A, B and C in above formula. So,
[tex]A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}[/tex]
Comparing the powers both sides,
2n+m=1 ...(1)
2m-n=1 ...(2)
Now, solving equation (1) and (2) we get :
[tex]n=\dfrac{1}{5}\\\\m=\dfrac{3}{5}[/tex]
Hence, the correct option is (E).
The values of the exponents n and m are [tex]\frac{1}{5}[/tex] and [tex]\frac{3}{5}[/tex] respectively.
The given parameters;
[tex]A = B^nC^m[/tex]
[tex]A = LT\\\\B = L^2T^{-1}\\\\C = LT^2[/tex]
The values of the given exponents "n" and "m" are calculated as follows;
[tex]LT = [L^2T^{-1}]^n[LT^2]^m\\\\LT= [L^{2n}T^{-n}][L^mT^{2m}]\\\\LT = [L^{2n + m} \ T^{2m-n}]\\\\L^1 = L^{2n + m} \\\\T^1 = T^{2m-n}\\\\1 = 2n \ + m \ ---(1)\\\\1 = 2m - n \ ---(2)[/tex]
from equation(2);
[tex]n = 2m - 1[/tex]
substitute the value of n into equation (1);
[tex]1 = 2(2m-1) + m\\\\1 = 4m - 2 + m\\\\1 = 5m - 2\\\\3 = 5m\\\\m = \frac{3}{5} \\\\n = 2m - 1\\\\n = 2(\frac{3}{5} ) - 1\\\\n = \frac{6}{5} - 1 \\\\n = \frac{1}{5}[/tex]
Thus, the values of the exponents n and m are [tex]\frac{1}{5}[/tex] and [tex]\frac{3}{5}[/tex] respectively.
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An electric motor is used to operate a Carnot refrigerator with an interior temperature of 0.00 ◦C. Liquid water at 0.00 ◦C is placed into the refrigerator and transformed to ice at 0.00 ◦C. If the room temperature is 300. K, what mass of ice can be produced in one day by a 0.50 hp motor that is running continuously? Assume that the refrigerator is perfectly insulated and operates at the maximum theoretical efficiency.
Answer:
732492 g
Explanation:
Given that
Room Temperature = 300 K
Cold T = 0°C = 0 + 273 =273 K
Work done = 0.375 hp
ΔH of fusion of water is 6008 J per mol
The first thing we do is to Convert the power into J/s
Given that 1 hp = 746 W (J/s), then
0.375 hp = 0.375 * 746 W/hp
0.375 = 280 J/s
Then we find the heat per unit time
Q = (cold T / hot T - cold T ) x power
Q = [273 / (300 - 273)] * 280
Q =273/27 * 280
Q = 10.1
Q = 2831 J/s
Moles of ice per second = Q/ ΔH fusion
Moles of ice per second = 2831/6008 = Moles of ice per second = 0.471 mol /s
If we convert 1 day into second, we have
1 day = 1 day * ( 24 hours/day) * ( 60 minutes/hour) * (60 seconds/minutes)
1 * 24 * 3600 = 86400 second
moles of ice /day
0.471 mol / s ) x 86400 s / day =
40694 mols of ice
Molar mass is 18 g/mol
Mass of ice
40694 * 18 = 732492 g
Mass of ice produced per day is 732492 g
Olivia is rolling down a hill on a skateboard and is accelerating at 2.4 m/s2. After 3 seconds, she runs into Reid at the bottom of the hill. If her initial velocity was 1.5 m/s, what was her velocity when she hit Reid?
did you ever get the answer?
Answer:
8.7
Explanation:
A=2.4m/s2 1.5m/s + (2.4m/s2)(3s)
Vi=1.5m/s 1.5m/s + 7.2
Vf=? Vf= 8.7m/s
T=3s
An electron and a proton are separated by a distance of 1.6×10−10m
(roughly the diameter of a single atom). The masses of the electron and proton are mp=1.673×10−27kg
and me=9.11×10−31kg,
respectively. The elementary charge e=1.602×10−19C.
The universal gravitational constant G=6.67×10−11N⋅m2/kg2
and the coulomb constant k=8.988×109N⋅m2/C2
What is the magnitude Fe of the electric force between the electron and the proton? Fe= 7.11×10−9N
What is the magnitude Fg
of the gravitational force between the electron and the proton?
Fg= 31.37×10−49N
In this scenario, how many times stronger is the electric force than the gravitational force?
Answer:
Fe = 9 x 10⁻⁹ N
Fg = 3.97 x 10⁻⁴⁶ N
Fe = 2.26 x 10³⁷ Fg
Explanation:
First we find electric force by Coulomb's Law as follows:
Fe = kq₁q₂/r²
where,
Fe = electric force = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = q₂ = charges on electron and proton = 1.6 x 10⁻¹⁹ C
r = distance between electron and proton = 1.6 x 10⁻¹⁰ m
Therefore,
Fe = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(1.6 x 10⁻¹⁰ m)²
Fe = 9 x 10⁻⁹ N
Now we find gravitational force by Newton's Law of Gravitation as follows:
Fg = Gm₁m₂/r²
where,
Fg = gravitational force = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = mass of electron = 9.11 x 10⁻³¹ kg
m₂ = mass of proton = 1.673 x 10⁻²⁷ kg
r = distance between electron and proton = 1.6 x 10⁻¹⁰ m
Therefore,
Fg = (6.67 x 10⁻¹¹ N.m²/kg²)(9.11 x 10⁻³¹ kg)(1.673 x 10⁻²⁷ kg)/(1.6 x 10⁻¹⁰ m)²
Fg = 3.97 x 10⁻⁴⁶ N
Dividing both forces:
Fe/Fg = (9 x 10⁻⁹ N)/(3.97 x 10⁻⁴⁶ N)
Fe = 2.26 x 10³⁷ Fg
The electric force is 2.3 × 10^39 times greater than the gravitational force.
The gravitational force between the proton and the electron is obtained from;
Fg = Gmemp/r^2
Where;
me = mass of the electron
mp = mass of the proton
r = distance of separation
Fg = 6.67 × 10^−11 × 9.11 × 10^−31 × 1.673 × 10^−27/(1.6 × 10^−10)^2
Fg = 3.94 × 10^−48 N
For the electric force;
Fe= K e^2/r^2
K = electric constant
e = magnitude of charge on electron and proton
r = distance of separation
Hence;
Fe = 8.988 × 10^9 × (1.602×10^−19)^2/(1.6×10^−10)^2 =
Fe = 9 × 10^−9 N
The number of time the electric force is stronger than the gravitational force = 9 × 10^−9 N/3.94 × 10^−48 N = 2.3 × 10^39 times
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The Moon's center is 3.9x10 m from Earth's center. The Moon is 1.5x10^8 km from the Sun's center. If the mass of the Moon is 7.3x10^22 kg, find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon
Explanation:
It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.
The gravitational force is given by :
[tex]F=\dfrac{Gm_em_m}{r^2}[/tex]
It means [tex]F\propto \dfrac{1}{r^2}[/tex]
So,
[tex]\dfrac{F_1}{F_2}=\dfrac{r_2}{r_1}[/tex]
r₁ = 3.9x10⁸ km
r₂= 1.5x10⁸ km
So,
[tex]\dfrac{F_1}{F_2}=\dfrac{1.5\times 10^8}{3.9\times 10^8}\\\\\dfrac{F_1}{F_2}=\dfrac{5}{13}[/tex]
Hence, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is 5:13.
Which is developed during the process of technology design
Answer:
solution
Explanation:
A(n) __________ is developed during the process of technological design.
Helppppppppppppppppppppppppppp
Answer:
Car 2 is travelling in a much higher speed than Car 1.But they are travelling or meeting in the same acceleration
A car moves in a straight line at a speed of 55.4 km/h.
How long (in seconds) will it take the car to move 1.21 km at this speed?
55.4 km => 1hr (60 min) (60×60=120 sec)
1 km => 120 ÷ 55.4
1.21 km => (120 ÷ 55.4) × 1.21
= 2.6209 sec ( 5sf )
= 2.62 sec ( 3sf )
i hope you are able to understand my solution :))
Describe how excavations are done.
Answer:
It includes earthwork, trenching, wall shafts, tunneling and underground
Explanation:
A ball is tossed up into the air with an initial speed of 5.0. How long does it take to return to the person's hand
Answer:
Explanation:
[tex]y(t) = 5t-5t^2[/tex]
When y(t) = 0, the ball is on the hand.
[tex]0=5t-5t^2\\0=5t(1-t)\\t=0,1[/tex]
It takes 1 second.
There are several different possibilities.
==> If the 5.0 means 5 miles per hour, that's 2.24 meters per second, up.
The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (2.24/9.8) = 0.229 second. Then, after is starts to fall, it takes the same amount of time to the person's hand.
Total time = 0.457 second.
==> If the 5.0 means 5 meters per second, up . . .
The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (5.0/9.8) = 0.51 second. Then, after is starts to fall, it takes the same amount of time to the person's hand.
Total time = 1.02 second.
==> If the 5.0 means 5 km/minute, that's about 83.33 meters per second, up.
The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (83.33/9.8) = 8.503 seconds. Then, after is starts to fall, it takes the same amount of time to the person's hand.
Total time = 17.01 seconds.
==> If the 5.0 means 5 furlongs per fortnight, that's about 0.00083 meters per second, up.
The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (0.00083/9.8) = 0.000085 second. Then, after is starts to fall, it takes the same amount of time to the person's hand.
Total time = 0.00017 second.
This is why all of your numbers always need their units.
What can make your computer "sick"?
Answer:
An virus
Explanation:
An computer virus can come from websites that try to steal your Information or apps that's not necessarily scanned and protected so that yor computer can be safe from viruses.
A computer may encounter problems or get "ill" for a variety of reasons. The system can become infected with malware and viruses, which can cause performance issues and data loss.
A failing hard drive or defective RAM are examples of hardware issues that can lead to instability and crashes. Overheating can lead to hardware damage or performance deterioration and is frequently brought on by insufficient cooling or clogged air vents.
Systems may become unstable as a result of software conflicts between out-of-date or incompatible programmes.
Electrical problems or power surges can harm components and interfere with proper operation. Computer issues can also be caused by human error, such as inadvertent file deletion or poor configuration.
Thus, a computer can be kept in good condition with routine maintenance, security precautions, and appropriate usage.
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Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is at the origin. The third charge, q3 = - 30 μC, is located at x = 2.0 m. What is the force on q2?
(a) 1.65 N in the negative x- direction
(b) 3.15 N in the positive x- direction
(c) 1.50 N in the negative x- direction
(d) 4.80 N in the positive x- direction
(e) 4.65 N in the negative x- direction.
Answer:
3.15 N towards the positive x-axis
Explanation:
first charge has charge q1 = 10 μC = 10 x 10^-6 C
second charge has charge q2 = 20 μC = 20 x 10^-6 C
third charge has charge q3 = -30 μC = -30 x 20^-6 C
According to coulomb's law, force between two charged particle is given as
F = [tex]\frac{-kQq}{r^2}[/tex]
Where
F is the force between the charges
k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.
Q is the magnitude of one charge
q is the magnitude of the other charge
is the distance between these two charges
For the force on q2 due to q1,
distance r between them = 0 - (-1.0) = 1 m
F = [tex]\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}[/tex] = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)
For the force on q2 due to q3,
distance between them = 2.0 - 0 = 2 m
F = [tex]\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}[/tex] = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)
Resultant force on q2 = 1.8 N + 1.35 N = 3.15 N towards the positive x-axis
Excess electrons are placed on a small lead sphere with a mass of 7.70 g so that its net charge is −3.35 × 10^−9 C.A) Find the number of excess electrons on the sphere.
B) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol?
Answer:
a
[tex]N = 2.094*10^{10} \ electrons[/tex]
b
[tex]O = 9.33*10^{-13} \ electrons[/tex]
Explanation:
From the question we are told that
The mass of the lead sphere is [tex]m = 7.70g = 0.0077 \ kg[/tex]
The net charge is [tex]Q_{net} = -3.35*10^{-9} \ C[/tex]
The atomic number is [tex]u = 82[/tex]
The molar mass is [tex]M = 207 \ g/mol[/tex]
Generally the excess number of electron on the sphere is mathematically represented as
[tex]N = \frac{Q_{net}}{ e }[/tex]
Here e is the charge on the electron is [tex]e = -1.60 *10^{-19} \ C[/tex]
So
[tex]N = \frac{-3.35 *10^{-19}}{ -1.60*10^{-19}}[/tex]
[tex]N = 2.094*10^{10} \ electrons[/tex]
Generally the number of atom present is mathematically represented as
[tex]n = N_a * \frac{m}{ M}[/tex]
Here [tex]N_a[/tex] is the Avogadro's number with value [tex]N_a = 6.0*10^{23} \ atoms[/tex]
[tex]n = 6.03 *10^{23} * \frac{7.70}{ 207}[/tex]
[tex]n = 2.24 *10^{22} \ atoms [/tex]
Generally the electrons are there per lead atom is mathematically represented as
[tex]O = \frac{N}{n}[/tex]
=> [tex]O = \frac{2.24*10^{22}}{2.094*10^{10}}[/tex]
=> [tex]O = 9.33*10^{-13} \ electrons[/tex]
A parking lot is going to be 60 m wide and 240 m long. what dimensions could be used for a scale model of the lot?
Answer:
it is A or 20x80 cm
Explanation: DID IT ON APEX
WOO
A girl is standing in front of a magic mirror. She finds the image of her head bigger the middle portion of her body of the same size and that of the legs smaller. The order of combinations for the magic mirror from the top is
Two violin strings are tuned to the same frequency 294 H. The tension in one string is then decreased by 2.0%. What will be the beat frequency heard when the two strings are played together?
Answer:
Beat frequency together = 2.95 Hz (Approx)
Explanation:
Given:
Frequency (F) = 294 H
Decrease in tension = 2%
Find:
Beat frequency together
Computation:
Tension = (100 - 2) / 100
Tension (T) = 0.98
Beat frequency together = Frequency (F) - (√T × F)
Beat frequency together = 294 - (√0.98 × 294)
Beat frequency together = 2.95 Hz (Approx)
The beat frequency heard when the two strings are played together is 2.95 Hz.
Given data:
The tuning frequency of the violin is, f = 294 Hz.
Decrement in the tension is, 2 %.
Since, tension is reduced at the rate of 2%. Then the new magnitude of tension on the string is,
T = (100 - 2 )/100
T = 0.98
Then the expression for the beat frequency heard when the two strings are played together is given as,
[tex]f_{b}=f -(\sqrt{T \times f})[/tex]
Solving as,
[tex]f_{b}=294-(\sqrt{0.98 \times 294})\\\\f_{b}=2.95\;\rm Hz[/tex]
Thus, we can conclude that the beat frequency heard when the two strings are played together is 2.95 Hz.
Learn more about the beat frequency here:
https://brainly.com/question/20347530
fter a great many contacts with the charged ball, how is any charge on the rod arranged (when the charged ball is far away)? Select the best description. View Available Hint(s) Select the best description. positive charge on end B and negative charge on end A negative charge spread evenly on both negative charge on end A with end B remaining neutral both ends neutral positive charge spread evenly on both
Answer:
With this analysis the correct answer is
Metal bar. The answer is uniform charge distribute
Non-metallic bar. As we see the charge of the ball, at the ends closest to the ball a charge of the same sign as the ball and at the far end the charge of equal magnitude and opposite sign
Explanation:
When the ball touches the bar, a charge transfer process occurs, leaving half in the bar and the other half in the ball, when the ball bounces and clears it, two dependent processes can occur, the bar is metallic or non-metallic , let's analyze the two cases
* metal bar, the load that is on the bar is distributed evenly throughout the bar, with each bounce of the ball the load on the bar increases and decreases in the balls until the two loads are equal and appear from this rebound no there is more load transfer
* Non-conductive insulating bar, when the ball touches the bar for the first time, it creates a load equal to q / 2, when the ball moves away from the bar, for this non-conductive, the transferred load remains in the same place. Therefore this load induces a load of equal magnitude, but opposite sign at the other end of the bar.
When the ball approaches for the second time, the opening already has a load, therefore the amount of load deposited is less, with between processes in each bounce the bar of the bar sounds, the process continues until the load of the ball and the bar have been equal.
With this analysis the correct answer is
Metal bar. The answer is uniform charge distribute
Non-metallic bar. As we see the charge of the ball, at the ends closest to the ball a charge of the same sign as the ball and at the far end the charge of equal magnitude and opposite sign
You leave your home at 1pm. At 3pm, you are 100 km east of your house. What was your average velocity in km/hr
Answer:
50km/h
Explanation:
Average velocity = change in distance (or distance travelled) divided by/ the change in time (or time taken.)
The change in distance has been given as 100km.
The change in time is 3pm-1pm = 2 hours.
Therefore the average velocity was 100/2 = 50km/h (to the east).
Hope this helped!
How is speed shown on a distance-time graph?
as the x value at the beginning of the line
as the x value at the end of the line
as the length of the line
as the slope of the line
Answer:
Divide the change in distance by the change in time.
Explanation: Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).
To my best knowledge It's D
_____ is the bending of waves as they pass by a barrier or through an opening.
A. Interference
B. Coherence
C. Diffraction
D. None of the above
Answer:
C. Diffraction
Explanation:
Diffraction is the bending of waves as they pass by a barrier or through an opening. One of the condition for diffraction is that the size of the aperture or obstacle is comparable to the wavelength of light. Fresnel diffraction and Fraunhofer diffraction are two types of diffraction.
Hence, the correct option is (c) "Diffraction".